chapter 3 probability distributions · 2017. 7. 13. · 50 . chapter 3 probability distributions ....

50

CHAPTER 3 PROBABILITY DISTRIBUTIONS Page Contents 31 Introduction to Probability Distributions 51 32 The Normal Distribution 56 33 The Binomial Distribution 60 34 The Poisson Distribution 64 Exercise 68 Objectives After working through this chapter you should be able to (i) understand basic concepts of probability distributions such as

random variables and mathematical expectations (ii) show how the Normal probability density function may be used to

represent certain types of continuous phenomena (iii) demonstrate how certain types of discrete data can be represented by

particular kinds of mathematical models for instance the Binomial and Poisson probability distributions

Chapter 3 Probability Distributions

51

31 Introduction to Probability Distributions

311 Random Variables

A random variable (RV) is a variable that takes on different numerical values determined by the outcome of a random experiment

Example 1

An experiment of tossing a coin 4 times

Notation Capital letter X - Random variable Lowercase x - a possible value of X

A random variable is discrete if it can take on only a limited number of values

A random variable is continuous if it can take any value in an interval

The probability distribution of a random variable is a representation of the probabilities for all the possible outcomes This representation might be algebraic graphical or tabular

A table or a formula listing all possible values that a discrete variable can take on together with the associated probability is called a discrete probability distribution

Example 2

The probability distribution of the number of heads when a coin is tossed 4 times

x 0 1 2 3 4 Pr(X = x) 1

16 4

16 6

16 4

16 1

16


52

ie Pr(X = x) =

4

16x

x = 0 1 2 3 4

In graphic form

1 Total area of rectangle = 1 2 Pr(X = 1) = shaded area

Example 3 An experiment of tossing two fair dice Let random variable X be the sum of two dice

The probability distribution of X Sum x 2 3 4 5 6 7 8 9 10 11 12 P(X = x) 1

36 2

36 3

36 4

36 5

36 6

36 5

36 4

36 3

36 2

36 1

36

The probability function f(x) of a discrete random variable X expresses the probability that X takes the value x as a function of x That is ( ) ( )Prf x X x= = where the function is evaluated at all possible values of x Properties of probability function ( )Pr X x= -

1 ( )Pr 0X x= ge for any value x 2 The individual probabilities sum to 1 that is ( )Pr 1

xX x= =sum

Example 4

Find the probability function of the number of boys on a committee of 3 selected at random from 4 boys and 3 girls


53

Continuous Probability Distribution 1 The total area under this curve bounded by the x axis is equal to one 2 The area under the curve between lines x = a and x = b gives the probability

that X lies between a and b which can be denoted by Pr(a le X le b) 3 We call f(x) a probability density function ie pdf

312 Mathematical Expectations Expectations for Discrete Random variables The expected value is the mean of a random variable Example 5

A review of textbooks in a segment of the business area found that 81 of all pages of text were error-free 17 of all pages contained one error while the remaining 2 contained two errors Find the expected number of errors per page

Let random variable X be the number of errors in a page

x ( )Pr X x= 0 081 1 017 2 002


54

Expected number of errors per page = 0times081 + 1times017 + 2times002 = 021 The expected value [ ]E X of a discrete random variable X is defined as

[ ]E X or ( )PrX

xx X xmicro = =sum

Definition Let X be a random variable The expectation of the squared discrepancy about the mean ( )2

XE X micro minus is called the variance denoted 2Xσ and given by

( ) ( )

( ) ( )

( )

22

2

2 2

or

Pr

Pr

X X

Xx

Xx

Var X E X

x X x

x X x

σ micro

micro

micro

= minus

= minus =

= = minus

sum

sum

Properties of a random variable

Let X be a random variable with mean Xmicro and variance 2Xσ and a b are constants

1 [ ] XE aX b a bmicro+ = +

2 ( ) 2 2XVar aX b a σ+ =

Sums and Differences of random variables

Let X and Y be a pair of random variables with means Xmicro and Ymicro and variances 2

Xσ and 2

Yσ and a b are constants

1 [ ] X YE aX bY a bmicro micro+ = +

2 [ ] X YE aX bY a bmicro microminus = minus 3 If X and Y are independent random variables then

( ) 2 2 2 2X YVar aX bY a bσ σ+ = +

( ) 2 2 2 2

X YVar aX bY a bσ σminus = +


55

Measurement of risk Standard Deviation Example 6 PROJECT A PROJECT B Profit(x) Pr(X=x) xPr(X=x) Profit(x) Pr(X=x) xPr(X=x) 150 03 45 (400) 02 (80) 200 03 60 300 06 180 250 04 100 400 01 40 800 01 80 Expected value = 205 Expected value = 220 === ===

Without considering risk choose B But Variance (X) = ( ) Pr( )x X xminus =sum micro 2 there4 Variance (A) = (150 minus 205)2(03) + (200 minus 205)2(03) + (250 minus 205)2(04) = 1725 SD(A) = 4153

Variance (B) = (minus400 minus 220)2(02) + (300 minus 220)2(06) + (400 minus 220)2(01)

+ (800 minus 220)2(01) = 117600 SD(B) = 34293

there4 Risk averse management might prefer A

Coefficient of Variation (CV)

Risk can be compared more satisfactorily by taking the ratio of the standard deviation to the mean of profit That is

CV = Standard deviation 100Mean

times

there4 CV of project A = 4153

205 100times

= 203


56

CV of project B = 342 93220

100times

= 1559 As a result B is more risky

32 The Normal Distribution Definition A continuous random variable X is defined to be a normal random variable if its probability function is given by f x x( )

( )exp[ ( ) ]= minus

minus12

12

2

σ πmicro

σ for minusinfin lt x lt +infin

where micro = the mean of X σ = the standard deviation of X π asymp 314159 Example 7 The following figure shows three normal probability distributions each of which has the same mean but a different standard deviation Even though these curves differ in appearance all three are ldquonormal curvesrdquo


57

Notation X ~ N(micro σ2) Properties of the normal distribution- 1 It is a continuous distribution 2 The curve is symmetric and bell-shaped about a vertical axis through the mean micro ie

mean = mode = median = micro 3 The total area under the curve and above the horizontal axis is equal to 1 4 Area under the normal curve Approximately 68 of the values in a normally distributed population within 1

standard deviation from the mean Approximately 955 of the values in a normally distributed population within 2


standard deviation from the mean Definition The distribution of a normal random variable with micro = 0 and σ = 1 is called a standard normal distribution Usually a standard normal random variable is denoted by Z Notation Z ~ N(0 1)


58

Remark Usually a table of Z is set up to find the probability P(Z ge z) for z ge 0 Example 8 Given Z ~ N(0 1) find (a) P(Z gt 173) (b) P(0 lt Z lt 173) (c) P(minus242 lt Z lt 08) (d) P(18 lt Z lt 28) (e) the value z that has (i) 5 of the area below it (ii) 3944 of the area between 0 and z Theorem If X is a normal random variable with mean micro and standard deviation σ then

XZ microσminus

=

is a standard normal random variable and hence

( ) 1 21 2Pr Pr x xx X x Zmicro micro

σ σminus minus lt lt = lt lt

Example 9 Given X ~ N(50 102) find P(45 lt X lt 62)


59

Example 10 The charge account at a certain department store is approximately normally distributed with an average balance of $80 and a standard deviation of $30 What is the probability that a charge account randomly selected has a balance (a) over $125 (b) between $65 and $95 Solution Let X be the balance of charge account ($) X ~ N(80 302) Example 11 On an examination the average grade was 74 and the standard deviation was 7 If 12 of the class are given As and the grades are curved to follow a normal distribution what is the lowest possible A and the highest possible B Solution Let X be the examination grade X ~ N(74 72)


60

33 The Binomial Distribution A binomial experiment possesses the following properties 1 There are n identical observations or trials 2 Each trial has two possible outcomes one called ldquosuccessrdquo and the other ldquofailurerdquo

The outcomes are mutually exclusive and collectively exhaustive for each trial 3 The probabilities of success p and of failure 1 minus p remain the same for all trials 4 The outcomes of trials are independent of each other Example 12 1 In testing 10 items as they come off an assembly line where each test or trial may

indicate a defective or a non-defective item 2 Five cards are drawn with replacement from an ordinary deck and each trial is

labelled a success or failure depending on whether the card is red or black Definition In a binomial experiment with a constant probability p of success at each trial the probability distribution of the binomial random variable X the number of successes in n independent trials is called the binomial distribution Notation X ~ b(n p)

P(X = x) = nx

p qx n x minus x = 0 1 hellip n

p + q = 1 Example 13 Of a large number of mass-produced articles one-tenth are defective Find the probabilities that a random sample of 20 will obtain (a) exactly two defective articles (b) at least two defective articles Solution Let X be the number of defective articles in the 20 X ~ b(20 01)


61

Example 14 A test consists of 6 questions and to pass the test a student has to answer at least 4 questions correctly Each question has three possible answers of which only one is correct If a student guesses on each question what is the probability that the student will pass the test Solution Let X be the number of correctly answered questions in the 6 X ~ b(6 13) Theorem The mean and variance of the binomial distribution with parameters n and p are micro = np and σ2 = npq respectively where p + q = 1


62

Example 15 A packaging machine produces 20 percent defective packages A random sample of ten packages is selected what are the mean and standard deviation of the binomial distribution of that process Solution Let X be the number of defective packages in the 10 X ~ b(10 02) The Normal Approximation to the Binomial Distribution Theorem Given X is a random variable which follows the binomial distribution with parameters n and p then P X x P x np

npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05

if n is large and p is not close to 0 or 1 (ie 01 lt p lt 09) Remark If both np and nq are greater than 5 the approximation will be good


63

Example 16 A process yields 10 defective items If 100 items are randomly selected from the process what is the probability that the number of defective items exceeds 13 Solution Let X be the number of defective items in the 100 X ~ b(100 01) Example 17 A multiple-choice quiz has 200 questions each with four possible answers of which only one is the correct answer What is the probability that sheer guesswork yields from 25 to 30 correct answers for 80 of the 200 problems about which the student has no knowledge Solution Let X be the number of correct answers in the 80 X ~ b(80 frac14)


64

34 The Poisson Distribution Experiments yielding numerical values of a random variable X the number of successes (observations) occurring during a given time interval (or in a specified region) are often called Poisson experiments A Poisson experiment has the following properties 1 The number of successes in any interval is independent of the number of successes in

other non-overlapping intervals 2 The probability of a single success occurring during a short interval is proportional

to the length of the time interval and does not depend on the number of successes occurring outside this time interval

3 The probability of more than one success in a very small interval is negligible Examples of random variables following Poisson Distribution 1 The number of customers who arrive during a time period of length t 2 The number of telephone calls per hour received by an office 3 The number of typing errors per page 4 The number of accidents per day at a junction Definition The probability distribution of the Poisson random variable X is called the Poisson distribution


65

Notation X ~ Po(λ) where λ is the average number of successes occuring in the given

time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip

e asymp 2718282 Example 18 The average number of radioactive particles passing through a counter during 1 millisecond in a laboratory experiment is 4 What is the probability that 6 particles enter the counter in a given millisecond Solution Let X be the number of radioactive particles passing through the counter in 1 millisecond X ~ Po(4) Example 19 Ships arrive in a harbour at a mean rate of two per hour Suppose that this situation can be described by a Poisson distribution Find the probabilities for a 30-minute period that (a) No ships arrive (b) Three ships arrive


66

Solution Let X be the number of arrivals in 30 minutes X ~ Po(1) Theorem The mean and variance of the Poisson distribution are both equal to λ Poisson approximation to the binomial distribution If n is large and p is near 0 or near 100 in the binomial distribution then the binomial distribution can be approximated by the Poisson distribution with parameter λ = np Example 20 If the probability that an individual suffers a bad reaction from a certain injection is 0001 determine the probability that out of 2000 individuals more than 2 individuals will suffer a bad reaction According to binomial Required probability

= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+

Using Poisson distribution

Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus

Required probability = 2

51e

minus = 0323

General speaking the Poisson distribution will provide a good approximation to binomial when (i) n is at least 20 and p is at most 005 or (ii) n is at least 100 the approximation will generally be excellent provided p lt 01 Example 21 Two percent of the output of a machine is defective A lot of 300 pieces will be produced Determine the probability that exactly four pieces will be defective Solution Let X be the number of defective pieces in the 300 X ~ b(300 002)


68

EXERCISE PROBABILITY DISTRIBUTIONS 1 If a set of measurements are normally distributed what percentage of these differ

from the mean by (a) more than half the standard deviation (b) less than three quarters of the standard deviation 2 If x is the mean and s is the standard deviation of a set of measurements which are

normally distributed what percentage of the measurements are (a) within the range ( )x splusmn 2 (b) outside the range ( )x splusmn 12 (c) greater than ( )x sminus 15 3 In the preceding problem find the constant a such that the percentage of the cases (a) within the range ( )x asplusmn is 75 (b) less than ( )x asminus is 22 4 The mean inside diameter of a sample of 200 washers produced by a machine is

502mm and the Std deviation is 005mm The purpose for which these washers are intended allows a maximum tolerance in the diameter of 496 to 508mm otherwise the washers are considered defective Determine the percentage of defective washers produced by the machine assuming the diameters are normally distributed

5 The average monthly earnings of a group of 10000 unskilled engineering workers

employed by firms in northeast China in 1997 was Y1000 and the standard deviation was Y200 Assuming that the earnings were normally distributed find how many workers earned

(a) less than Y1000 (b) more than Y600 but less than Y800 (c) more than Y1000 but less than Y1200 (d) above Y1200 6 If a set of grades on a statistics examination are approximately normally distributed

with a mean of 74 and a standard deviation of 79 find (a) The lowest passing grade if the lowest 10 of the students are give Fs (b) The highest B if the top 5 of the students are given As


69

7 The average life of a certain type of a small motor is 10 years with a standard deviation of 2 years The manufacturer replaces free all motors that fail while under guarantee If he is willing to replace only 3 of the motors that fail how long a guarantee should he offer Assume that the lives of the motors follow a normal distribution

8 Find the probability that in a family of 4 children there will be (a) at least 1 boy (b) at

least 1 boy and 1 girl Assume that the probability of a male birth is 12

9 A basketball player hits on 75 of his shots from the free-throw line What is the

probability that he makes exactly 2 of his next 4 free shots 10 A pheasant hunter brings down 75 of the birds he shoots at What is the

probability that at least 3 of the next 5 pheasants shot at will escape If X represents the number of pheasants that escape when 5 pheasants are shot at find the probability distribution of X

11 A basketball player hits on 60 of his shots from the floor What is the probability

that he makes less than one half of his next 100 shots 12 A fair coin is tossed 400 times Use the normal-curve approximation to find the

probability of obtaining (a) Between 185 and 210 heads inclusive (b) Exactly 205 heads (c) Less than 176 or more than 227 heads 13 Ten percent of the tools produced in a certain manufacturing process turn out to be

defective Find the probability that in a sample of 10 tools chosen at random exactly two will be defective by using

(a) the binomial distribution (b) the Poisson approximation to the binomial 14 Suppose that on the average 1 person in every 1000 is an alcoholic Find the

probability that a random sample of 8000 people will yield fewer than 7 alcoholics 15 Suppose that on the average 1 person in 1000 makes a numerical error in preparing

his income tax return If 10000 forms are selected at random and examined find the probability that 6 7 or 8 of the forms will be in error


70

16 A secretary makes 2 errors per page on the average What is the probability that she makes

(a) 4 or more errors on the next page (b) no error 17 The probability that a person dies from a certain respiratory infection is 0002 Find

the probability that fewer than 5 of the next 2000 so infected will die

CHAPTER 3 PROBABILITY DISTRIBUTIONS



312 Mathematical Expectations

32 The Normal Distribution

33 The Binomial Distribution

34 The Poisson Distribution

EXERCISE PROBABILITY DISTRIBUTIONS


51



A random variable (RV) is a variable that takes on different numerical values determined by the outcome of a random experiment

Example 1

An experiment of tossing a coin 4 times

Notation Capital letter X - Random variable Lowercase x - a possible value of X

A random variable is discrete if it can take on only a limited number of values

A random variable is continuous if it can take any value in an interval

The probability distribution of a random variable is a representation of the probabilities for all the possible outcomes This representation might be algebraic graphical or tabular

A table or a formula listing all possible values that a discrete variable can take on together with the associated probability is called a discrete probability distribution

Example 2

The probability distribution of the number of heads when a coin is tossed 4 times

x 0 1 2 3 4 Pr(X = x) 1

16 4

16 6

16 4

16 1

16


52

ie Pr(X = x) =

4

16x

x = 0 1 2 3 4

In graphic form




36 2

36 3

36 4

36 5

36 6

36 5

36 4

36 3

36 2

36 1

36



xX x= =sum

Example 4



53






x ( )Pr X x= 0 081 1 017 2 002


54


[ ]E X or ( )PrX

xx X xmicro = =sum



( ) ( )

( ) ( )

( )

22

2

2 2

or

Pr

Pr

X X

Xx

Xx

Var X E X

x X x

x X x

σ micro

micro

micro

= minus

= minus =

= = minus

sum

sum




2 ( ) 2 2XVar aX b a σ+ =



Xσ and 2




( ) 2 2 2 2X YVar aX bY a bσ σ+ = +

( ) 2 2 2 2



55




+ (800 minus 220)2(01) = 117600 SD(B) = 34293





times


205 100times

= 203


56


100times




minus12

12

2

σ πmicro




57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













52

ie Pr(X = x) =

4

16x

x = 0 1 2 3 4

In graphic form




36 2

36 3

36 4

36 5

36 6

36 5

36 4

36 3

36 2

36 1

36



xX x= =sum

Example 4



53






x ( )Pr X x= 0 081 1 017 2 002


54


[ ]E X or ( )PrX

xx X xmicro = =sum



( ) ( )

( ) ( )

( )

22

2

2 2

or

Pr

Pr

X X

Xx

Xx

Var X E X

x X x

x X x

σ micro

micro

micro

= minus

= minus =

= = minus

sum

sum




2 ( ) 2 2XVar aX b a σ+ =



Xσ and 2




( ) 2 2 2 2X YVar aX bY a bσ σ+ = +

( ) 2 2 2 2



55




+ (800 minus 220)2(01) = 117600 SD(B) = 34293





times


205 100times

= 203


56


100times




minus12

12

2

σ πmicro




57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













53






x ( )Pr X x= 0 081 1 017 2 002


54


[ ]E X or ( )PrX

xx X xmicro = =sum



( ) ( )

( ) ( )

( )

22

2

2 2

or

Pr

Pr

X X

Xx

Xx

Var X E X

x X x

x X x

σ micro

micro

micro

= minus

= minus =

= = minus

sum

sum




2 ( ) 2 2XVar aX b a σ+ =



Xσ and 2




( ) 2 2 2 2X YVar aX bY a bσ σ+ = +

( ) 2 2 2 2



55




+ (800 minus 220)2(01) = 117600 SD(B) = 34293





times


205 100times

= 203


56


100times




minus12

12

2

σ πmicro




57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













54


[ ]E X or ( )PrX

xx X xmicro = =sum



( ) ( )

( ) ( )

( )

22

2

2 2

or

Pr

Pr

X X

Xx

Xx

Var X E X

x X x

x X x

σ micro

micro

micro

= minus

= minus =

= = minus

sum

sum




2 ( ) 2 2XVar aX b a σ+ =



Xσ and 2




( ) 2 2 2 2X YVar aX bY a bσ σ+ = +

( ) 2 2 2 2



55




+ (800 minus 220)2(01) = 117600 SD(B) = 34293





times


205 100times

= 203


56


100times




minus12

12

2

σ πmicro




57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













55




+ (800 minus 220)2(01) = 117600 SD(B) = 34293





times


205 100times

= 203


56


100times




minus12

12

2

σ πmicro




57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













56


100times




minus12

12

2

σ πmicro




57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













57







58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













58


XZ microσminus

=






59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













59



60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













60





P(X = x) = nx




61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













61



62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













62


npqZ x np

npq( ) ( ( )

( )( )

( ))= =

minus minuslt lt

+ minus05 05



63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













63



64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













64






65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













65


time interval

P(X = x) = ex

xminusλλ

x = 0 1 2 hellip



66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













66


= ( ) ( ) ( ) ( ) ( ) ( )12000

00 001 0 999

20001

0 001 0 9992000

20 001 0 9990 2000 1 1999 2 1998minus

+

+


Pr(0 suffers) = 2

20 10

2e

e=

minus

λ = np = 2

Pr(1 suffers) = 2

21 21

2e

e=

minus


67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













67

Pr(2 suffer) = 2

22 22

2e

e=

minus


51e

minus = 0323



68










69















70













68










69















70













69















70













70












chapter 3 probability distributions · 2017. 7. 13. · 50 . chapter 3 probability distributions ....

Documents